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I am not quite sure the whether following statement is correct:

If $A\in \sigma(A_1,\cdots, A_n)$ and $A$ is independent of $A_1$, then $A\in \sigma(A_2,\cdots, A_n)$. I can see that the sets in $\sigma(A_1,\cdots, A_n)$ which are independent of $A$ form a $\sigma$ algebra. But I just cannot prove it is indeed $\sigma(A_2,\cdots, A_n)$.

It looks quite reasonable, since intuitively $A$ is in $\sigma(A_2,\cdots, A_n)$ means $A$ can only be affected by $A_2,\cdots, A_n$. So I guess this should be true.

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  • $\begingroup$ Independence requires a probabilty measure. But the conclusion has nothing to do with measures. So the question hardly makes sense. $\endgroup$
    – Kavi Rama Murthy
    Commented Oct 8, 2020 at 9:35
  • $\begingroup$ @KaviRamaMurthy Thanks for comment. I see, this makes sense. $\endgroup$
    – Zorualyh
    Commented Oct 8, 2020 at 9:55

1 Answer 1

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Let $P(A)=0$ but $A \neq \emptyset$. Let $n=2, A_2=\emptyset$ and $A=A_1$. This is a counter-example.

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